Author 
Topic: Pythagorean triples (Read 1430 times) 

Christine
Full Member
Posts: 159


Pythagorean triples
« on: Jun 7^{th}, 2014, 12:20pm » 
Quote Modify

Pythagorean triples (a, b, c) then a^2 + b^2 = c^2 How to prove that c^2  a*b and c^2 + a*b can both be expressed as a sum of two squares?


IP Logged 



0.999...
Full Member
Gender:
Posts: 156


Re: Pythagorean triples
« Reply #1 on: Jun 8^{th}, 2014, 5:49am » 
Quote Modify

c^2ab=( 1/2(c+(ba)) )^2 + ( 1/2(c(ba)) )^2 and c^2+ab=( 1/2(c+(b+a)) )^2 + ( 1/2(c(b+a)) )^2 I really only examined two cases and extrapolated. Perhaps someone else will be able to offer more insight.


IP Logged 



rloginunix
Uberpuzzler
Posts: 1026


Re: Pythagorean triples
« Reply #2 on: Jun 10^{th}, 2014, 8:14pm » 
Quote Modify

The search for all the positive integer solutions of the algebraic equation x^2 + y^2 = z^2 has the geometric equivalent of finding all the right triangles with the integral side lengths. If I remember correctly early on the Babylonians raised that question which also interested the Greek geometers of the time. Pythagoras himself is credited with the following solution: a = 2n + 1, b = 2n^2 + 2n, c = 2n^2 + 2n + 1 for his a^2 + b^2 = c^2 triple (n is an arbitrary positive integer). My idea leads to lots of writing but is simple: a*b = (2n + 1)(2n^2 + 2n) = 4n^3 + 4n^2 + 2n^2 + 2n = 4n^3 + 6n^2 + 2n c^2 = (2n^2 + 2n + 1)^2 = 4n^4 + 4n^2(2n + 1) + 4n^2 + 4n + 1 = 4n^4 + 8n^3 + 8n^2 + 4n + 1 Rearrangement of your difference, for example: c^2  a*b = 4n^4 + 8n^3 + 8n^2 + 4n + 1  4n^3  6n^2  2n = 4n^4 + 4n^3 + 2n^2 + 2n + 1 = 4n^4 + 4n^3 + n^2 + (n^2 + 2n + 1) = = 4n^4 + 4n^3 + n^2 + (n + 1)^2 = n^2(4n^2 + 4n + 1) + (n + 1)^2 = n^2(2n + 1)^2 + (n + 1)^2 = = (n(2n + 1))^2 + (n + 1)^2 You can rename the terms "n(2n + 1)" and "n + 1" as some other integers of your liking and you have your sum of two squares. Perhaps now you can try your hand at "c^2 + a*b".


IP Logged 



0.999...
Full Member
Gender:
Posts: 156


Re: Pythagorean triples
« Reply #3 on: Jun 11^{th}, 2014, 5:03am » 
Quote Modify

rloginlinux, while it is an infinite class of solutions, the parametrization does not cover all primitive solutions. An example of one that is missed is (8,15,17). In the textbook on Number Theory by Ireland and Rosen, excercise 1.23, one shows that the following twovariable paramatrization, covers all primitive pythagorean triples (a,b,c) up to a permutation of a and b: a = 2uv, b = v^2u^2, c = v^2+u^2. Using this parametrization, the calculations are rather nice: c^2ab=v^4+u^4+2u^2v^22uv^3+2u^3v=v^42uv^3+u^2v^2+u^4+2u^3v+u^2v^2=v^2(vu)^2+u^2(v+u)^2. The calculation in the case c^2+ab is the same, except with u and v reversed.


IP Logged 



Immanuel_Bonfils
Junior Member
Posts: 114


Re: Pythagorean triples
« Reply #4 on: Jul 30^{th}, 2014, 8:14pm » 
Quote Modify

on Jun 8^{th}, 2014, 5:49am, 0.999... wrote:c^2ab=( 1/2(c+(ba)) )^2 + ( 1/2(c(ba)) )^2 and c^2+ab=( 1/2(c+(b+a)) )^2 + ( 1/2(c(b+a)) )^2 
 Are you sure about this?


IP Logged 



